Question

A salt solution that contains 0.90 g of NaCl in 100.0 mL of solution.

Answer 1

Do you want a particular concentration unit? Common results are below (I assume ideal behavior and, where needed, density ≈ 1.00 g·mL–1).

Given: 0.90 g NaCl in 100.0 mL solution.
Molar mass NaCl = 58.44 g·mol–1 → moles NaCl = 0.90 / 58.44 = 0.01541 mol.

- Molarity (M) = moles / L = 0.01541 / 0.1000 = 0.1541 M ≈ 0.15 M (2 s.f.).
- % (w/v) = 0.90 g per 100 mL = 0.90% (w/v).
- % (w/w) (if density ≈1.00 g·mL–1 so mass solution = 100.0 g): 0.90 g / 100.0 g = 0.90% (w/w).
- Molality (m) (assuming mass solvent = 100.0 g − 0.90 g = 99.10 g = 0.09910 kg): 0.01541 / 0.09910 = 0.1555 m ≈ 0.16 m.
- Osmolarity (assuming full dissociation, i ≈ 2): 2 × 0.1541 = 0.3082 Osm·L–1 ≈ 0.31 Osm·L–1 (≈ 310 mOsm·L–1).

Tell me which of these you need, or if you want results with a measured density or different assumptions.